C ++内存地址分配

Question

Apologies i got frustrated and posted question through mobile without proper details

Consider the following C++ code:

 int arr[2];
    arr[0] = 10;
    arr[1] = 20;
   // cout << &arr;
    for (int i = 0; i < 2; i++)
    {
       cout << &arr + i << "\t\t"<<endl;
    }

    cout << sizeof (arr);

cout in for loop prints following

0x7ffeefbff580 0x7ffeefbff588

which is 8 bytes farther than the first element

My question is why it is 8 bytes further and not 4 bytes if on my machine sizeof(int) is 4?

Answer 1

Now that you gave us the code we can answer your question.

So the confusing piece is this: &arr + i . This does not do what you think it does. Remember that & takes precedence over + . And so you take address of arr and move it forward i times.

Pointer arithmetic works in such a way that &x + 1 moves the pointer forward by size(x) . So in your case what is size(arr) ? It is 8 because it is 2-element array of integers (I'm assuming ints are of size 4). And so &arr + 1 actually moves the pointer 8 bytes forward. The exact thing you experience. You don't ask for next int, you ask for next array . I encourage you to play around, for example define arr as int[3] (which is of size 12) and see how the pointer moves 12 bytes forward.

So first solution is to do arr + i without & . We can apply pointer arithmetic to an array in which case it decays to a pointer type int* . Now since int is of size 4 then arr + 1 points correctly to a memory segment 4 bytes forward.

But what I suggest is to stay away from pointer arithmetic and do &arr[i] instead. This does the same thing but IMO is less error prone, less confusing and tells us more about the intent.